Two charges +8q and -2q are fixed on X-axis at origin and `x=+a` locations. A third charge +q is to be located on X-axis (other than infinitely far away) so that it is in equilibrium. The location of the third charge is correctly represented by

Options

A. `x=2a`
B. `x=3a//2`
C. `x=-a`
D. `x=3a`

Explanation

To find the position of the third charge +q such that it is in equilibrium, we can follow these steps:

### Step 1: Understand the Configuration
We have two fixed charges:
- Charge 1: +8q at the origin (x = 0)
- Charge 2: -2q at x = +a

We need to place a third charge +q on the x-axis such that the net force acting on it is zero.

### Step 2: Determine Possible Regions for Charge +q
Since +q is a positive charge, it will experience a repulsive force from +8q and an attractive force from -2q. We need to analyze the possible positions for +q:
1. To the left of +8q (x < 0)
2. Between +8q and -2q (0 < x < a)
3. To the right of -2q (x > a)

### Step 3: Analyze Each Region
- **Left of +8q (x < 0)**: The force from +8q will push +q to the right (repulsion), and the force from -2q will pull +q to the left (attraction). This will not lead to equilibrium as both forces will act in the same direction.

- **Between +8q and -2q (0 < x < a)**: Here, +q will be repelled by +8q and attracted by -2q. The net force will not be zero as the repulsion will be stronger due to the proximity to +8q.

- **Right of -2q (x > a)**: In this region, +q will be repelled by -2q and attracted by +8q. There is a possibility for equilibrium here.

### Step 4: Set Up the Force Equations
Let’s assume the position of +q is at a distance r from the origin (x = 0). The distance from +8q to +q is r, and the distance from -2q to +q is (r - a).

The forces acting on +q due to +8q and -2q can be expressed as:
- Force due to +8q (F1):
\[ F_1 = k \cdot \frac{8q \cdot q}{r^2} \]
- Force due to -2q (F2):
\[ F_2 = k \cdot \frac{2q \cdot q}{(r - a)^2} \]

### Step 5: Set the Forces Equal for Equilibrium
For equilibrium, the magnitudes of the forces must be equal:
\[ F_1 = F_2 \]
This gives us:
\[ k \cdot \frac{8q^2}{r^2} = k \cdot \frac{2q^2}{(r - a)^2} \]
Cancelling out common terms (k and q^2):
\[ \frac{8}{r^2} = \frac{2}{(r - a)^2} \]

### Step 6: Cross Multiply and Solve
Cross multiplying gives:
\[ 8(r - a)^2 = 2r^2 \]
Expanding and simplifying:
\[ 8(r^2 - 2ar + a^2) = 2r^2 \]
\[ 8r^2 - 16ar + 8a^2 = 2r^2 \]
\[ 6r^2 - 16ar + 8a^2 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 6, b = -16a, c = 8a^2 \):
\[ r = \frac{16a \pm \sqrt{(-16a)^2 - 4 \cdot 6 \cdot 8a^2}}{2 \cdot 6} \]
\[ r = \frac{16a \pm \sqrt{256a^2 - 192a^2}}{12} \]
\[ r = \frac{16a \pm \sqrt{64a^2}}{12} \]
\[ r = \frac{16a \pm 8a}{12} \]
This gives two solutions:
1. \( r = \frac{24a}{12} = 2a \)
2. \( r = \frac{8a}{12} = \frac{2a}{3} \) (not valid as it is between the two charges)

### Conclusion
The only valid position for the charge +q in equilibrium is at \( r = 2a \).

Thus, the location of the third charge is correctly represented by **2a**.

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